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a, vì 13=1x13=(-1)x(-13) mà x,y thuộc Z =>x-8 ;y+2 thuộc Z nên ta có bảng sau:
x-8 1 13 -1 -13
y+2 13 1 -13 -1
x 9 21 7 -5
y 11 -1 -15 -3
vậy (x;y) thuộc {(9;11);(21;-1);(7;-15);(-5;-3)}
b,vì 11=1x11=(-1)x(-11) mà x;y thuộc Z =>x-2 ;y+7 thuộc Z nên ta có bảng sau:
x-2 1 11 -1 -11
y+7 11 1 -11 -1
x 3 13 1 -9
y 4 -6 -18 -8
vậy (x;y) thuộc {(3;4);(13;-6);(1;-18);(-9;-8}
a) Vì (x-8)(y+2)=13 suy ra:
X-8;y+2 thuộc ước của 13 suy ra:
x-8;y+2 thuộc 13;-13;1;-1
Ta có bảng :
x+8. 13 -13 1 -1
y+2. 1 -1 13 -13
x. 21 -5 9 7
y. -1 -3 11 -15
Vậy : x=... ;y=...
![](https://rs.olm.vn/images/avt/0.png?1311)
xy=-11, x<y
x,y nguyên => \(\hept{\begin{cases}x=-11\\y=1\end{cases}}\)
Vậy (x;y)=(-11;1)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{8}{9}\) : ( 2 - 3 \(\times\) y) = \(\dfrac{5}{3}\)
2 - 3 \(\times\) y = \(\dfrac{8}{9}\) : \(\dfrac{5}{3}\)
2 - 3 \(\times\) y = \(\dfrac{8}{15}\)
3 \(\times\) y = 2 - \(\dfrac{8}{15}\)
3 \(\times\) y = \(\dfrac{22}{15}\)
y = \(\dfrac{22}{15}\) : 3
y = \(\dfrac{22}{45}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(3x+2xy=7\)
\(\Leftrightarrow x\left(2y+3\right)=7\)
\(\Leftrightarrow x;2y+3\inƯ\left(7\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\2y+3=7\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\2y+3=-7\end{matrix}\right.\\\left\{{}\begin{matrix}x=7\\2y+3=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-7\\2y+3=-1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=-5\end{matrix}\right.\\\left\{{}\begin{matrix}x=7\\y=-\dfrac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-7\\y=-2\end{matrix}\right.\end{matrix}\right.\)
Vậy ...
b/ \(3x-5xy=11\)
\(\Leftrightarrow x\left(3-5y\right)\inƯ\left(11\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\3-5y=11\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\3-5y=-11\end{matrix}\right.\\\left\{{}\begin{matrix}x=11\\3-5y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-11\\3-5y=-1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=-\dfrac{8}{5}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=\dfrac{14}{5}\end{matrix}\right.\\\left\{{}\begin{matrix}x=7\\y=\dfrac{2}{5}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-7\\y=-\dfrac{4}{5}\end{matrix}\right.\end{matrix}\right.\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x,y ={ -1,11}
b) x,y= { -1,15; -15,1; -3,5; -5,3}
mình chỉ biết đến vậy tui
xin lỗi
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 9:
Ta có: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{z}{-17}=\dfrac{-t}{-9}\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{-z}{17}=\dfrac{t}{9}=-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-2\\\dfrac{-y}{3}=-2\\\dfrac{-z}{17}=-2\\\dfrac{t}{9}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\-y=-6\\-z=-34\\t=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=6\\z=34\\t=-18\end{matrix}\right.\)
Vậy: (x,y,z,t)=(-10;6;34;-18)
Bài 11:
Ta có: \(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)
\(\Leftrightarrow\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}=\dfrac{-7}{6}\)
Ta có: \(\dfrac{x}{18}=\dfrac{-7}{6}\)
\(\Leftrightarrow x=\dfrac{18\cdot\left(-7\right)}{6}=-21\)
Ta có: \(\dfrac{-98}{y}=\dfrac{-7}{6}\)
\(\Leftrightarrow y=\dfrac{-98\cdot6}{-7}=84\)
Ta có: \(\dfrac{-14}{z}=\dfrac{-7}{6}\)
\(\Leftrightarrow z=\dfrac{-14\cdot6}{-7}=12\)
Ta có: \(\dfrac{u}{-78}=\dfrac{-7}{6}\)
\(\Leftrightarrow u=\dfrac{-78\cdot\left(-7\right)}{6}=\dfrac{78\cdot7}{6}=91\)
Ta có: \(\dfrac{t}{102}=\dfrac{-7}{6}\)
\(\Leftrightarrow t=\dfrac{-7\cdot102}{6}=-7\cdot17=-119\)
Vậy: (x,y,z,t,u)=(-21;84;12;-119;91)
2/y=11/66
<=>2/y=1/6
<=>y=2:1/6
<=>y=2.6
<=>y-12
Vậy y=12
Y x 11 = 2 x 66 => Y = 2 x 66 / 11 = 12